|Aug24-12, 05:00 PM||#205|
Bell's Paradox Question
Originally Posted by ghwellsjr
If the force is applied only at the front, that creates the maximum expansive stress possible without applying reverse thrust to the rear. SO adding a forward thrust at the rear actually reduces the overall expansive stress, so I am confused as to why you think this would lead to expansive disruption where the single thrust would not.
The basic premise of Born rigid acceleration; that stressless acceleration would necessitate a scaled and distributed acceleration scheme is certainly reasonable.
But aside from the fact that it is unrealistic in application , until we develop some totally new science that negates inertia and momentum,(gravity drive or???) stress is an inevitable consequence of acceleration, and stress, per se, is not a big problem. We live every day under a constant stress of 1 g. The relevant concern is if that stress is constant or dynamically increasing. So why do you think it would be increasing to the point of disruption?
You also did not provide any basis for your assumption that equal thrust at the front and rear would necessarily result in equal coordinate acceleration at those points and a constant separation in the launch frame.
|Aug24-12, 06:06 PM||#206|
|Aug28-12, 11:44 AM||#207|
austinO post 191
U is the universal rest frame. A and B space ships pass U at t=0, moving at v = .6c.
Both experience equal length contraction to .8L in the x direction. If length contraction
is a result of em deformation in response to acceleration, then length expansion should
be the response to deceleration. If the A ship returns to U and stops, it should recover
its original length.
According to SR, if A moves away from B, B should measure a length contraction of A.
At first it seems A would have to expand and contract simultaneously to satisfy both
requirements, but not so. The confusion occurs because there are two different length
contractions, the first due to absolute motion relative to light speed, the second due
to perception. Since U is the absolute rest frame, the A & B contraction is the result of
em phenomena. After deceleration of A to v=0, it expands to 1L. Now consider B as
passing A at rest in the U frame. Time dilation for B is .8t, thus B arrives early at
locations on the Ux axis. Since everything in the B frame slows B trusts his clock and
interprets the effects as length contraction of the U frame, thus A is contracted to
.8L, and both requirements are met.
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